Question

A concave mirror have a focal lenght of 20 cm, At what distance from the mirror should a object of 4 cm be placed so that it forms a image at a distance of 30 cm from the mirror ? Also calculate the sise of the image formed.

plzz.. help me​

Answer 1

Explanation:

Given:-

• Focal length = 20cm

• Height of object = 4cm

• Image distance = 30cm

To Find:-

• Object distance = ?

• Size of the image ( Height of image)

Solution:-

$\sf For \: a \: concave \: mirror:-$

$\sf Height \: of \: object \: h_{o} = +ve$

$\sf Height \: of \: image \: h_{i} = -ve$

$\sf Object \: distance \: (u) = -ve$

$\sf Image \: distance \: (v) = -ve$

$\sf Now, \: According \: to \: the mirror \: formula$

$\sf \dashrightarrow \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\sf \implies \dfrac{1}{-30} + \dfrac{1}{u} = \dfrac{1}{-20}$

$\sf \implies \dfrac{1}{u} = \dfrac{1}{-20} + \dfrac{1}{30}$

$\sf \implies \dfrac{1}{u} = \dfrac{-3+ 2}{60}$

$\sf \implies \dfrac{1}{u} = \dfrac{-1}{60}$

$\sf \implies u = -60$

$\sf Now,\: Magnification = \dfrac{-v}{u} =\dfrac{h_{i}}{h_{o}}$

$\sf \implies \dfrac{-v}{u} =\dfrac{h_{i}}{h_{o}}$

$\sf \implies \dfrac{-(-30)}{-60} =\dfrac{h_{i}}{4}$

$\sf \implies \dfrac{1}{-2} =\dfrac{h_{i}}{4}$

$\sf \implies h_{i}= \dfrac{4}{-2}$

$\sf\implies h_{i}= -2$

$\underline{\boxed{\pink{\rm \therefore Size\: of \: image = 2cm}}}$

$\underline{\boxed{\purple{\rm Object\: should \: be \: kept \: 60m \: infront\: of \: the\: mirror}}}$